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3.4.2 1 sample variance

1 sample variance Purpose of the 1 sample variance test The 1 sample variance test is used to determine whether the variability of a single population is equal to a specified target (or known) variance. - Question answered: “Is the process variance equal to a specific value?” - Typical use: Compare current process variability to: - A design requirement - A historical variance - A customer specification for spread This test is based on the chi-square distribution and is applied to continuous, normally distributed data. --- Core concepts: variance and standard deviation Variance and standard deviation definitions Variance and standard deviation describe data spread around the mean. - Variance (σ² or s²): - Population variance: average squared distance of each value from the population mean - Sample variance: estimate of population variance from sample data - Standard deviation (σ or s): - Square root of variance - Expressed in the same units as the data The 1 sample variance test is directly performed on variance, but in practice you often start from the sample standard deviation. Population vs. sample variance - Population variance (σ²): - Parameter of the entire process or population - Unknown in most real settings - Sample variance (s²): - Calculated from a finite sample - Used as an estimator of σ² The 1 sample variance test uses s² from your sample to make an inference about σ². --- Assumptions of the 1 sample variance test Correct interpretation of the test requires specific conditions. - Normality: - The underlying population must be normally distributed - The chi-square approach is sensitive to non-normality - Independence: - Sample observations are independent of each other - Single population: - Data come from one consistent process or population When these assumptions are significantly violated, the chi-square test for variance may give misleading results. --- Hypotheses for the 1 sample variance test Null and alternative hypotheses You compare the unknown population variance σ² to a hypothesized variance σ₀². - Null hypothesis (H₀): σ² = σ₀² (The true variance equals the target or historical variance.) - Alternative hypothesis (H₁) can take three forms: - Two-sided: σ² ≠ σ₀² - Upper-sided: σ² > σ₀² - Lower-sided: σ² < σ₀² Choice of one-sided vs. two-sided depends on the practical question. Typical practical formulations - Concerned about increased variability: - H₀: σ² = σ₀² - H₁: σ² > σ₀² - Concerned about reduced variability (e.g., risk of over-tight control): - H₀: σ² = σ₀² - H₁: σ² < σ₀² - Concerned about any change in variability: - H₀: σ² = σ₀² - H₁: σ² ≠ σ₀² --- Test statistic and chi-square distribution Test statistic formula Given a sample of size n with variance s², and a hypothesized population variance σ₀², the test statistic is: - Chi-square statistic: χ² = (n − 1)·s² / σ₀² where: - n = sample size - s² = sample variance - σ₀² = hypothesized variance under H₀ Under the null hypothesis and normality, χ² follows a chi-square distribution with (n − 1) degrees of freedom. Degrees of freedom - Degrees of freedom (df): n − 1 - The chi-square distribution shape depends strongly on df: - For small df, it is skewed to the right - As df increases, it becomes more symmetric Correct df is essential for accurate critical values and p-values. --- Critical regions and decision rules Two-sided test For a significance level α: - Critical values: - Lower critical value: χ²(L) = χ²(α/2, df) - Upper critical value: χ²(U) = χ²(1 − α/2, df) - Decision rule (two-sided): - Reject H₀ if χ² ≤ χ²(L) or χ² ≥ χ²(U) - Fail to reject H₀ otherwise This checks for variance either significantly smaller or larger than σ₀². One-sided upper test For H₁: σ² > σ₀²: - Critical value: - Upper critical: χ²(U) = χ²(1 − α, df) - Decision rule: - Reject H₀ if χ² ≥ χ²(U) - Fail to reject H₀ otherwise This focuses only on detecting an increase in variance. One-sided lower test For H₁: σ² < σ₀²: - Critical value: - Lower critical: χ²(L) = χ²(α, df) - Decision rule: - Reject H₀ if χ² ≤ χ²(L) - Fail to reject H₀ otherwise This focuses only on detecting a decrease in variance. --- p-values for the 1 sample variance test Two-sided p-value For observed test statistic χ²(obs) and df: - If χ²(obs) > expected center (df): - p-value = 2 × [1 − F(χ²(obs))] - If χ²(obs) < expected center: - p-value = 2 × F(χ²(obs)) where F is the cumulative distribution function of the chi-square distribution with df degrees of freedom. One-sided p-values - Upper-sided (H₁: σ² > σ₀²): - p-value = P(χ² ≥ χ²(obs)) = 1 − F(χ²(obs)) - Lower-sided (H₁: σ² < σ₀²): - p-value = P(χ² ≤ χ²(obs)) = F(χ²(obs)) Interpretation: - Small p-value (≤ α): evidence against H₀ in favor of H₁ - Large p-value (> α): insufficient evidence to reject H₀ --- Confidence interval for the population variance Interval for variance A (1 − α)·100% confidence interval for σ² is: - Lower limit: (n − 1)·s² / χ²(1 − α/2, df) - Upper limit: (n − 1)·s² / χ²(α/2, df) where χ²(p, df) is the pth quantile of the chi-square distribution. Interval for standard deviation If needed, take the square root of each limit: - Lower limit for σ: √(lower variance limit) - Upper limit for σ: √(upper variance limit) The confidence interval provides a plausible range for the true variance (or standard deviation), not just a yes/no test outcome. --- Practical applications and interpretation When to use the 1 sample variance test Use this test when: - You have a single sample from a process - A target or historical variance (or standard deviation) is specified - You want to assess whether current variability matches that value Examples include checks on: - Process spread relative to a known historical level - Equipment performance compared with a specified variance - Measurement system variability relative to an acceptable limit (when modeled as a single variance comparison) Interpreting results for improvement - Reject H₀, σ² > σ₀²: - Evidence that the process is more variable than intended - Suggests need to: - Investigate special causes - Reduce sources of variation - Reject H₀, σ² < σ₀²: - Evidence that the process is less variable than the hypothesized value - Could support claims of improved consistency - Fail to reject H₀: - No strong statistical evidence of a variance difference from σ₀² - Does not prove equality, but suggests data are consistent with σ² = σ₀² In all cases, interpretation should consider: - Sample size and power - Assumption of normality - Practical significance, not only statistical significance --- Common pitfalls and checks Normality concerns The chi-square based 1 sample variance test assumes normality. - Use preliminary checks: - Histograms or probability plots - Knowledge of process behavior - If data are strongly non-normal: - Results of the chi-square variance test may be unreliable - Transformations or alternative methods may be necessary, but those go beyond this specific test Misuse of specification limits - The 1 sample variance test compares to a specific variance, not directly to specification limits. - Confusion to avoid: - Comparing variance to tolerance width without proper translation - If needed: - Convert specification-based expectations into a target variance or standard deviation and test against that value. --- Summary The 1 sample variance test assesses whether the variance of a single population matches a specified value using the chi-square distribution. It relies on a sample variance s², compares it to a hypothesized variance σ₀² through the statistic χ² = (n − 1)·s² / σ₀² with n − 1 degrees of freedom, and evaluates this against critical values or p-values under normality and independence assumptions. One- or two-sided hypotheses allow focus on increased, decreased, or any change in variability. Confidence intervals for the variance (and standard deviation) complement the test by providing a plausible range for the true variability. Proper application and interpretation require attention to assumptions, sample size, and the practical implications of differences in variance.

Practical Case: 1 sample variance A regional lab processes water-quality tests for a bottling plant. The plant’s contract requires that lab turnaround time (TAT) “must be consistently within 24 hours” so production planning is reliable. Recently, production managers complain that TAT feels “all over the place,” even though the average still appears close to 24 hours. The Quality Manager, a Lean Six Sigma Black Belt, suspects rising variability rather than a shift in the mean. She pulls one recent sample of 30 consecutive TAT records from the lab’s system and calculates the sample variance of turnaround time. She compares it to the variance level that had been typical and acceptable when the contract was last renewed. The new 1 sample variance is substantially higher than the historical benchmark, confirming that turnaround time has become more inconsistent, even though the average hasn’t changed much. Using this evidence, she: - Justifies a focused improvement project with the lab supervisor. - Maps the process and discovers batching at one approval step as the main source of fluctuation. - Implements a simple “no batch larger than 5 samples” rule and standard work. A follow-up 1 sample variance check one month later shows variance back near the original acceptable level, and production reports far fewer schedule disruptions. End section

Practice question: 1 sample variance A Black Belt is assessing output variability of a filling process. She collects one rational subgroup of n = 10 bottles and calculates a sample variance of 0.25 ml². Which of the following best describes the role of this 1-sample variance in her analysis? A. It is an unbiased estimate of the population variance, used to quantify long-term process center B. It is an unbiased estimate of the population variance, used to quantify short-term process spread C. It is a biased estimate of the population variance, used only for capability indices D. It is an unbiased estimate of the population standard deviation, used to quantify short-term process spread Answer: B Reason: The 1-sample variance s² estimated from a rational subgroup is an unbiased estimator of the population variance σ² and is used to quantify the process dispersion (short-term spread) for that characteristic. Other options are incorrect because it does not describe center (A), is not defined as biased in standard practice (C), and it estimates variance, not standard deviation (D). --- A Black Belt wants to test whether the variance of a critical dimension equals the historical value of 0.04 mm², assuming normality. He has a sample of size n = 16 and computes s² = 0.08 mm². Which is the appropriate hypothesis test? A. 1-sample t-test on the mean B. F-test comparing two sample variances C. Chi-square test for 1-sample variance D. 1-sample Z-test on the variance Answer: C Reason: For testing a single population variance against a known target variance (under normality), the correct procedure is the chi-square test for 1-sample variance, using (n−1)s²/σ₀² ~ χ². Other options are incorrect as they test means (A, D) or compare two variances from two samples (B). --- A Black Belt calculates the 1-sample variance of a key CTQ from n = 9 observations and obtains s² = 4.0 units². For a 95% confidence interval on the population variance, which degrees of freedom and chi-square quantiles must be used? A. df = 8, χ²(0.025,8) and χ²(0.975,8) B. df = 9, χ²(0.025,9) and χ²(0.975,9) C. df = 8, χ²(0.05,8) and χ²(0.95,8) D. df = 9, χ²(0.05,9) and χ²(0.95,9) Answer: A Reason: The 1-sample variance uses df = n−1 = 8. A 95% confidence interval for σ² uses the 2.5% and 97.5% chi-square quantiles with df = 8, i.e., χ²(0.025,8) and χ²(0.975,8). Other options use incorrect df (B, D) or quantiles corresponding to a 90% interval (C, D). --- During Measure phase, a Black Belt computes two different 1-sample variances from the same process: one using raw data from mixed shifts (s² = 10.0), and one using a rational subgroup from a single shift (s² = 2.5). What is the most appropriate interpretation? A. The process is non-normal, invalidating the 1-sample variance calculations B. The within-shift variance is lower; between-shift variation inflates the overall variance C. The smaller variance must be an error because variance cannot differ within the same process D. The larger variance must be used exclusively for control chart limits Answer: B Reason: A smaller 1-sample variance from a rational subgroup versus a larger variance from mixed conditions indicates that within-subgroup (short-term) variation is lower, while between-shift sources increase total variability. Other options incorrectly invoke non-normality without evidence (A), deny legitimate variance decomposition (C), or overstate that only the larger variance can be used in control charts (D). --- A Black Belt has 20 independent measurements of a dimension that is assumed normal. The sample variance is s² = 1.5 mm². She wants to test at α = 0.05 whether the population variance exceeds 1.0 mm². Which decision rule is correct? A. Reject H₀ if (n−1)s²/1.0 < χ²(0.95,19) B. Reject H₀ if (n−1)s²/1.0 > χ²(0.95,19) C. Reject H₀ if (n−1)s²/1.0 > χ²(0.975,19) D. Reject H₀ if (n−1)s²/1.0 < χ²(0.025,19) Answer: B Reason: For H₀: σ² ≤ 1.0 vs H₁: σ² > 1.0 (upper one-sided test, α = 0.05), reject H₀ if the chi-square statistic (n−1)s²/σ₀² exceeds χ²(0.95,19). Other options correspond to lower-tail rejection (A, D) or use the 0.975 quantile, which would give α = 0.025 in the upper tail (C).

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